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Let X be the space obtained from $S^2$ by identifying (x; y; 0) with (-x; -y; 0), for all (x; y; 0)$\in$ S2. Compute $\pi_(X).$.

I know to choose the open sets so that they each deformation retract to th real projective plane and the intersection is a circle, But what is the pushout? $Z_2*Z_2?$

Eddie
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1 Answers1

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It's been a while since I've made a calculation like this but here goes: As you suggest if you take enlargements of the northern and southern hemispheres you get two open sets $U_1, U_2$ which deformation retract onto a real projective plane and hence have fundamental groups isomorphic to $\mathbb{Z}_2$. Moreover their intersection deformation retracts onto a circle, which has fundamental group isomorphic to $\mathbb{Z}$. We also see that the maps induced $\mathbb{Z} \rightarrow \mathbb{Z}_2$ just send 1 to 1. Now in order to compute the fundamental group of our space (using Seifert-van Kampen) we take the free product $\pi_1(U_1) * \pi_1(U_2)$ and then quotient out by relations $i_1(x)\cdot i_2(x)^{-1}$ for $x \in \pi_1(U_1 \cap U_2)$. So this is just $\mathbb{Z}_2 * \mathbb{Z}_2 / <1\cdot 1'> \cong \mathbb{Z}_2$ .

Nate
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  • The equator is actually a circle with antipodal points identified, which is also a circle. So if you have a picture of a sphere in your head, the generator of the fundamental group for $\pi_1(U_1 \cap U_2)$ is not the whole equator, but rather just half of it (which becomes a loop when we quotient). This is also a generator for the fundamental group of the real projective place corresponding to either hemisphere, to the map sends 1 to 1. -- This was originally in response to someone's comment (I'm not crazy) – Nate Aug 27 '13 at 18:14